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Maths question

  1. Feb 10, 2005 #1

    VietDao29

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    Hi,
    I have been trying to solve it for some days,... but I couldnot find out. The equation is:
    [tex]a^{2} + b^{2} = c^{2}[/tex]
    a, b, c <> 0
    I try to call:
    [tex]x_{1} = \frac{b}{a}[/tex]
    [tex]x_{2} = \frac{c}{a}[/tex]
    And I have
    [tex]a^2 \times{(1 + x_{1}^{2} - x_{2}^{2})} = 0[/tex]
    And I solve for x1 and x2. But I didn't succeed.
    Can you suggest me some other ways??
    Thanks,
    Viet Dao,
     
  2. jcsd
  3. Feb 10, 2005 #2

    arildno

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    What are you after?
    Is it a formula for how pythogorean triplets in general can be found?
    Be more specific!
     
  4. Feb 10, 2005 #3

    dextercioby

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    If this must be the case,then:

    [tex] a=m^{2}-n^{2} [/tex]

    [tex] b=2mn [/tex]

    [tex] c=m^{2}+n^{2} [/tex]

    Where "m" and "n" are ARBITRARY.

    Daniel.
     
  5. Feb 10, 2005 #4

    VietDao29

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    I dunno really. My friend just asked me that some weeks ago, and I simply got stuck.
    I remember reading a book that says:
    [tex]a^{n} + b^{n} = c^{n}[/tex]
    Where a, b, c <> 0 and [itex]a, b, c \in Z[/itex]
    [itex]n \in N[/itex], n > 2. Then there will be no answer. But it didn't prove it.
    --------------
    Thanks Dextercioby, I got it now... But how can you know that?
    [tex] a=m^{2}-n^{2} [/tex]
    [tex] b=2mn [/tex]
    [tex] c=m^{2}+n^{2} [/tex]
    Is there another case?
    --------------
    Viet Dao,
     
  6. Feb 10, 2005 #5

    dextercioby

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    I didn't invent it,certainly...I read it somewhere,a long time ago.And it's not too hard to remember.

    No,that formula covers all possible "m" and "n" from R (in particular from N hence generating the pythagoreic triplets).

    Daniel.
     
  7. Feb 10, 2005 #6

    VietDao29

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    Hi,
    Thanks a lot dextercioby. :smile: That really helps.
    Viet Dao,
     
  8. Feb 10, 2005 #7

    Hurkyl

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    I think you need an additional factor to get all triples:

    a = p(m2 - n2)
    b = 2mnp
    c = p(m2 + n2)


    Of course, you don't need p for primitive triples.
     
  9. Feb 10, 2005 #8

    dextercioby

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    I see what u mean.Yes,the triplet (6,12,15) would not be found through "m" and "n"...

    Daniel.
     
  10. Feb 10, 2005 #9

    VietDao29

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    Hi,
    Thanks, Hurkyl. I appreciate your help.
    Viet Dao,
     
  11. Feb 10, 2005 #10

    dextercioby

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    Just a second.I gave the example with the triplet (6,12,15),which HAS SOLUTIONS,but for "m" and "n" in the reals.So my initial statement is still valid.

    Daniel.

    P.S.Yes,for all natural number triplets,generated by natural "m" and "n",one has to put the "p".
     
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